Growth and Decay Problems

Let N(t) denote amount of substance (or population) that is either growing or decaying. If we assume that `(dN)/(dt)` , the time rate of change of this amount of substance, is proportional to the amount of substance present, then `(dN)/(dt)=kN` where k is the constant of proportionality.

When k is positive, population grows, when k is negative - population decays.

We are assuming that N(t) is a differentiable, hence continuous, function of time. For population problems, where N(t) is actually discrete and integer-valued, this assumption is incorrect. Nonetheless, above model still provides a good approximation to she physical laws governing such a system.

This differential equation is separable. Rewriting it we obtain `(dN)/N=kdt` .

Integration of both sides gives: `ln(N)=kt+c_1` or `N=Ce^(kt)` where `C=e^(c_1)` .

If we are given that `N(0)=N_0` then`N_0=Ce^(k*0)` or `C=n_0` .

So, `N=N_0e^(kt)` .

Example 1. The population of a certain country is known to increase at a rate proportional to the number of people presently living in the country. If after two years the population has doubled, and after three years the population is 20000, estimate the number of people initially living in the country.

Let `N_0` is number of people initially living in the country, then `N(0)=N_0` , `N(2)=2N_0` , `N(3)=20000` .

So, as was shown above if `N(0)=N_0` then `N=N_0e^(kt)` .

Next, since `N(2)=2N_0` then `2N_0=N_0e^(k*2)` or `2k=ln(2)` . This gives that `k=ln(2)/2` .

Now, equation can be rewritten as `N=N_0e^(ln(2)/2t)` .

Since `N(3)=20000` then `20000=N_0e^(ln(2)/2 3)` or `N_0=20000e^(-1.5ln(2))~~7071` people.

Example 2. A certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour it is observed that 10 percent of the material has decayed, find the half-life (period of time it takes for the amount of material to decrease by half) of the material.